Assignment 8: Classification Trees
Jack Masella, Brendan Costello
2024-04-10
Objective
Predict customer attrition using customer credit card data
Data Prep
- Create a new variable called “Random” by classifying observations of “existing customer” and “attrited customer”
- Split the data (10127 Observations), into training and validation dataset (70%/30% , 7126/3001)
The Tree
Two Interpretations
- Customers who are 39 or older have the second highest number of customers who’ve become attrited
- Customers who’s transaction amount total is equal to or greater than 5448 are the least likely to become attrited
Variable Importance
Model Accuracy
Using the validation data, it is observed that the model is able to:
- Predicted correctly 83.15% of attrited customers
- Misclassify existing customers as attrited 3% of the time
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## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
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## Total Observations in Table: 3066
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##
## | validation_tree$Attrition_Flag_predicted
## validation_tree$Attrition_Flag | Attrited Customer | Existing Customer | Row Total |
## -------------------------------|-------------------|-------------------|-------------------|
## Attrited Customer | 383 | 98 | 481 |
## | 1456.529 | 240.172 | |
## | 0.796 | 0.204 | 0.157 |
## | 0.882 | 0.037 | |
## | 0.125 | 0.032 | |
## -------------------------------|-------------------|-------------------|-------------------|
## Existing Customer | 51 | 2534 | 2585 |
## | 271.021 | 44.690 | |
## | 0.020 | 0.980 | 0.843 |
## | 0.118 | 0.963 | |
## | 0.017 | 0.826 | |
## -------------------------------|-------------------|-------------------|-------------------|
## Column Total | 434 | 2632 | 3066 |
## | 0.142 | 0.858 | |
## -------------------------------|-------------------|-------------------|-------------------|
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## Statistics for All Table Factors
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## Pearson's Chi-squared test
## ------------------------------------------------------------
## Chi^2 = 2012.413 d.f. = 1 p = 0
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## Pearson's Chi-squared test with Yates' continuity correction
## ------------------------------------------------------------
## Chi^2 = 2006.027 d.f. = 1 p = 0
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